To solve this, we use the principles of conservation of angular momentum and energy considerations.

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Step 1: Information and Formulas

Given:

1. Rod mass ($M$): $0.950 \, \text{kg}$
2. Rod length ($L$): $1.20 \, \text{m}$
3. Ball mass ($m$): $0.265 \, \text{kg}$
4. Height from which the ball is dropped ($h_0$): $1.400 \, \text{m}$
5. Height the ball reaches after collision ($h_1$): $0.599 \, \text{m}$
6. Moment of inertia of the rod about its center: $I_\text{rod} = \frac{1}{12} M L^2$

To Find:

Angular velocity ($\omega$) of the rod after the collision.

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Step 2: Linear Momentum of the Ball Before and After Collision

The velocity of the ball before impact is calculated using: $v_\text{before} = \sqrt{2 g h_0}$ The velocity of the ball after impact is calculated using: $v_\text{after} = \sqrt{2 g h_1}$ Where $g = 9.8 \, \text{m/s}^2$.

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Step 3: Angular Momentum Conservation

The system's angular momentum about the pivot is conserved. Initially, only the ball has angular momentum since the rod is stationary.

Initial Angular Momentum (Before Collision):

$L_\text{initial} = m v_\text{before} \cdot \frac{L}{4}$ where $\frac{L}{4}$ is the perpendicular distance from the pivot to the point of impact.

Final Angular Momentum (After Collision):

After the collision, the rod rotates with angular velocity $\omega$, and the ball continues with linear velocity $v_\text{after}$. The total angular momentum is: $L_\text{final} = I_\text{rod} \cdot \omega + m v_\text{after} \cdot \frac{L}{4}$

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Step 4: Conservation Equation

Equating the initial and final angular momenta: $m v_\text{before} \cdot \frac{L}{4} = I_\text{rod} \cdot \omega + m v_\text{after} \cdot \frac{L}{4}$ Substitute $I_\text{rod} = \frac{1}{12} M L^2$ and solve for $\omega$: $\omega = \frac{m v_\text{before} \cdot \frac{L}{4} - m v_\text{after} \cdot \frac{L}{4}}{\frac{1}{12} M L^2}$

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Step 5: Substituting Values

1. $v_\text{before} = \sqrt{2 \cdot 9.8 \cdot 1.400} = 5.242 \, \text{m/s}$
2. $v_\text{after} = \sqrt{2 \cdot 9.8 \cdot 0.599} = 3.430 \, \text{m/s}$
3. $I_\text{rod} = \frac{1}{12} \cdot 0.950 \cdot (1.20)^2 = 0.114 \, \text{kg·m}^2$

Substitute into the equation: $\omega = \frac{0.265 \cdot (5.242 - 3.430) \cdot \frac{1.20}{4}}{0.114}$

Simplify: $\omega = \frac{0.265 \cdot 1.812 \cdot 0.30}{0.114} = \frac{0.144 \, \text{kg·m}^2/\text{s}}{0.114 \, \text{kg·m}^2} = 1.26 \, \text{rad/s}$

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Final Answer:

The angular velocity of the rod is: $\boxed{1.26 \, \text{rad/s}}$

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Let me know if you want any further clarification or additional steps. Here are some related questions to explore:

1. How does the height the ball reaches after collision affect the angular velocity of the rod?
2. What would happen if the ball hit the rod at a different point?
3. How would adding friction at the pivot change the results?
4. What if the rod had a different shape or moment of inertia?
5. Can energy conservation be used as an alternate method in this scenario?

Tip: Always check whether a problem involves angular or linear quantities to apply the correct conservation laws